[LeetCode] 13. Roman to Integer

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

• I can be placed before V (5) and X (10) to make 4 and 9. 
• X can be placed before L (50) and C (100) to make 40 and 90. 
• C can be placed before D (500) and M (1000) to make 400 and 900.

Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

罗马数字转换整数。

题意跟12题恰好相反。这个题依然没什么算法和思想可言，思路依然是需要创建一个整数和罗马数字之间的mapping，然后从右往左遍历input字符串。当遍历到 s[i] 的时候，需要看一下 s[i], cur 和 s[i - 1], pre 位置上的罗马数字哪个对应的整数大，分如下几种情况，用例子说明，

如果s = "III"，s[i] = s[i - 1] = "I"。此时pre == cur，只要把pre位置上代表的整数加到结果即可。这种case应该是只适用于 s[i] 是"I"的，因为没有其他字母可以这样重复出现，类似"XX"或者"VV"应该都是不合法的。

如果s = "IV"，s[i] = "V", s[i - 1] = "I"，此时pre < cur，需要把I的值从V减去（5-1 = 4）。

如果s = "VI"，s[i] = "I", s[i - 1] = "V"，此时pre >= cur，这种case同第一种case，需要将I的值和V的值相加（5 + 1 = 6）。

时间O(n)

空间O(n) - hashmap记录对应关系

JavaScript实现

/**
 * @param {string} s
 * @return {number}
 */
var romanToInt = function (s) {
    // corner case
    if (s === null || s.length === 0) {
        return 0;
    }

    // normal case
    const map = new Map([['I', 1], ['V', 5], ['X', 10], ['L', 50], ['C', 100], ['D', 500], ['M', 1000]]);
    let i = s.length - 1;
    let res = map.get(s[i]);
    while (i > 0) {
        let cur = map.get(s[i]);
        let pre = map.get(s[i - 1]);
        if (pre >= cur) {
            res += pre;
        } else {
            res -= pre;
        }
        i--;
    }
    return res;
};

Java实现

class Solution {
    public int romanToInt(String s) {
        int sum = 0;
        int preNum = getValue(s.charAt(0));
        for (int i = 1; i < s.length(); i++) {
            int num = getValue(s.charAt(i));
            if (preNum < num) {
                sum -= preNum;
            } else {
                sum += preNum;
            }
            preNum = num;
        }
        sum += preNum;
        return sum;
    }

    private int getValue(char ch) {
        switch (ch) {
        case 'I':
            return 1;
        case 'V':
            return 5;
        case 'X':
            return 10;
        case 'L':
            return 50;
        case 'C':
            return 100;
        case 'D':
            return 500;
        case 'M':
            return 1000;
        default:
            return 0;
        }
    }
}

LeetCode 题目总结